Combined Bank
Post: Senior Officer (IT)
Exam Date: 31/01/2022, Exam Taker: BSC
CREATE TABLE t (Val INT);
INSERT INTO t(val) VALUES (1,2,2,3,
"null", "null",4,5);
SELECT COUNT (val)
FROM t;
SELECT COUNT DISTICT (val)
FROM t;
CREATE TABLE:
A table t is created with one column val of type INTEGER.
INSERT INTO:
The values 1, 2, 2, 3, NULL, NULL, 4, 5 are inserted into the column val.
SELECT COUNT(val):
This query counts all non-NULL values in the val column.
Total non-NULL values: 1, 2, 2, 3, 4, 5
Result: 6
SELECT COUNT(DISTINCT val):
This query counts only the unique non-NULL values in the val column.
Unique values: 1, 2, 3, 4, 5
Result: 5
Note: Question Collect করা সম্ভব হয়নি।
i. Subnet Mask
ii. Block Size.
iii. Network Address
iv. Broadcast Address
v. Total valid Host.
Subnet Mask:/22 corresponds to the subnet mask 255.255.252.0.
Block Size: 2^(32−22)=2^10=1024.
Network Address: 172.16.16.0/22.
Broadcast Address: 172.16.19.255/22.
Total Valid Hosts: 2^(32−22)−2=1024−2=1022
Explanation
Given IP Address: 172.16.16.137/22
Step 1: /22 বোঝা
/22 মানে Network bits = 22
Host bits = 32 − 22 = 10 bits
Step 2: Subnet Mask লেখা
/22 এর Binary Subnet Mask:
11111111.11111111.11111100.00000000
Decimal Subnet Mask = 255.255.252.0
Step 3: IP Address কে Binary তে রূপান্তর
172 = 10101100
16 = 00010000
16 = 00010000
137 = 10001001
IP (Binary):
10101100.00010000.00010000.10001001
Step 4: AND Operation (Network Address)
IP Address AND Subnet Mask করা হয়।
যেখানে Subnet Mask = 0 → Host bits 0 হয়ে যায়।
IP: 10101100.00010000.00010000.10001001 Mask: 11111111.11111111.11111100.00000000 ------------------------------------------------ Result: 10101100.00010000.00010000.00000000
Binary Result কে Decimal এ রূপান্তর করলে:
Network Address = 172.16.16.0
Step 5: Broadcast Address বের করা
Broadcast address পেতে সব Host bits = 1 করা হয়।
Network (Binary): 10101100.00010000.00010000.00000000 Host bits (10 bits) = 1111111111 Broadcast (Binary): 10101100.00010000.00010011.11111111
Binary থেকে Decimal করলে:
Broadcast Address = 172.16.19.255
Step 6: Valid Host Range
First Host = Network + 1 → 172.16.16.1
Last Host = Broadcast − 1 → 172.16.19.254
Step 7: Total Valid Hosts
Host bits = 10
Total = 210 = 1024
Valid Hosts = 1024 − 2 = 1022
#include<stdio.h>
using namespace std;
// Recursive function to check if a string is a palindrome
bool isPalindrome(string str, int start, int end) {
// Base case: If the start index is greater than or equal to the end index
if (start >= end) {
return true;
}
// Check if the first and last characters are the same
if (str[start] != str[end]) {
return false;
}
// Recursive case: Check the substring excluding the first and last characters
return isPalindrome(str, start + 1, end - 1);
}
int main() {
string input;
cout << "Enter a string: "; cin >> input;
if (isPalindrome(input, 0, input.length() - 1)) {
cout << input << " is a palindrome." << endl;
} else {
cout << input << " is not a palindrome." << endl;
}
return 0;
}
Sample I/O:
Enter a string: noon
noon is a palindrome.
=== Code Execution Successful ===
